You, Archimedes, suspect that the king’s crown is not solid gold but is instead gold-plated lead. To test your theory, you weigh the crown, and find it to weigh 60.0 N, and to have an apparent weight of 56.4 N when it is completely submerged in water. Take the density of gold to be 19300 kg/m^3, the density of lead to be 11340 kg/m^3, and the density of water to be 1000 kg/m^3. Also, use g = 9.80 m/s^2. What percentage (by weight) of this crown is made of gold? The rest, by the way, is lead.
(a) What is the average density of the crown?
Correct: Your answer is correct. kg/m3
(b) What percentage (by weight) of this crown is made of gold? The rest, we assume, is lead.
Incorrect: Your answer is incorrect. %

Answer :

Answer:

a) 16675.75 Kg/m³ b) 77.6%

Explanation:

the weight of the crown = 60 N, density of gold = 19300 Kg/m^3, density of lead = 11340 kg/m^3, density of water = 1000kg/m^3 and acceleration due to gravity = 9.8 m/s^2

upthrust on the crown = weight in air - weight when fully submerged in water = 60 - 56.4 = 3.6 N

mass of water displaced = 3.6 / 9.8  since weight = mass × g

mass of water displaced = 0.367 Kg

density of water = mass / volume

1000 = 0.367 / volume

cross multiply and find volume

volume of the crown = 0.367 / 1000 = 0.000367 m³ since the crown will displace water of equal volume according to Archimedes principle

Let V1 represent the volume of Gold and let V2 represent the volume of lead

Total volume of the crown = V1 + V2

also

density of gold = mass of gold / V1 and density of lead = mass of lead / V2

19300 = mass of gold in the crown / V1 and 11340 = mass of lead in the crown / V2

19300 V1 = mass of gold and 11340 V2 = mass of lead

add the two together

19300 V1 + 11340 V2 = weigth of the crown / 9.8

19300 V1 + 11340 V2 = 6.12 also

V1 + V2 = 0.000367

make V1 subject of the formula in equation 2

V1 = 0.000367 - V2

substitute for V1 in equation 1

19300 (0.000367 - V2) + 11340 V2 = 6.12

open the bracket

7.083 - 19300 V2 + 11340 V2 = 6.12

rearrange the equation

-7960 V2 = 6.12 - 7.083  

-7960 V2 = -0.963

V2 = -0.963 / -7960 = 0.000121 (volume of lead in the crown)

substitute V2 into equation 2

V1 + 0.000121  = 0.000367m³

V1 = 0.000367 - 0.000121 = 0.000246m³ (volume of gold in the crown)

so mass of gold in the crown = 19300 × 0.000246 = 4.748 kg

and mass of lead = 11340 × 0.000121 = 1.372 kg

average density of the crown = (mass of gold + mass of lead) / total volume = 6.12 / 0.000367 = 16675.75 kg/ m³

b) percentage make of gold = mass of gold / total mass × 100 = 77.6 % approx

(a) The average density of the crown is 16,675.75  kg/m³

(b) The percentage of mass of gold in the crown is 77.6%

The given parameters;

  • weight of the crown in air, [tex]W_a[/tex] = 60 N
  • weight of the crown in water, [tex]W_w[/tex] = 56.4 N
  • density of gold, [tex]\rho_g[/tex] = 19,300 kg/m³
  • density of the lead, [tex]\rho _l = 11,340 \ kg/m^3[/tex]
  • density of water, [tex]\rho_w = 1000 \ kg/m^3[/tex]
  • acceleration due to gravity, g = 9.8 m/s²

The total mass of the crown is calculated as;

[tex]mass \ of \ crown = \frac{60 \ N}{9.8 \ m/s^2} = 6.12 \ kg[/tex]

Find the volume of the crown:

The upthrust of the crown = [tex]W_a - W_w = 60 - 56.4 = 3.6 \ N[/tex]

The volume of the water displaced = volume of the crown

The volume of the water displaced is calculated as follows;

[tex]volume = \frac{mass}{density} = \frac{m}{\rho_w} \\\\Upthrust = mg\\\\m = \frac{Upthrust}{g} = \frac{3.6}{9.8}= 0.367 \ kg[/tex]

[tex]Volume = \frac{0.367}{1000} = 0.000367 \ m^3[/tex]

The volume of the crown = 0.000367 m³

(a) The average density of the crown is calculated as;

[tex]average \ density \ of \ the \ crown = \frac{total \ mass \ of \ the \ crown}{Volume \ of \ the \ crown} \\\\average \ density \ of \ the \ crown = \frac{6.12 \ kg}{0.000367 \ m^3} \\\\[/tex]

[tex]average \ density \ of \ the \ crown = 16,675.75 \ kg/m^3[/tex]

Find the mass of gold in the crown:

The volume of the crown = volume of gold + volume of lead

[tex]0.000367 = \frac{mass \ of \ gold}{\rho_g} + \frac{mass \ of \ lead}{\rho _l} \\\\0.000367 = \frac{m_g}{19,300} + \frac{m_l}{11,340} \\\\(19,300 \times 11,340)(0.000367) = 11,340m_g \ \ + \ \ 19.300m_l\\\\80322.354= 11,340m_g \ \ + \ \ 19.300m_l \ \ -----(1)[/tex]

Also, the total mass of the gold and lead is given as;

[tex]m_g + m_l = \frac{60 \ N}{9.8} \\\\m_g + m_l = 6.12\\\\m_l = 6.12 - m_g ----(2)[/tex]

Substitute equation (2) in to (1) to solve for mass of the gold

[tex]80322.354 = 11,340 m_g + 19,300(6.12 - m_g)\\\\80322.354 = 11,340 m_g + 118116 - 19,300m_g\\\\80322.354 - 118116 = 11,340 m_g - 19,300m_g\\\\-37793.646 = -7,960 \ m_g\\\\m_g = \frac{37793.646}{7960} = 4.75 \ kg[/tex]

The mass of the gold = 4.75 kg

(b) The percentage of mass of gold in the crown is calculated as;

[tex]\% \ of \ Gold= \frac{mass \ of \ gold}{total \ mass} \times 100 \%\\\\\% \ of \ Gold= \frac{4.75}{6.12} \times 100 \%\\\\\% \ of \ Gold= 77.6\%[/tex]

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